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Lockpicking house rules for D&D

On the DM side, all you’ve got to do is add 8 to the DC listed in the module (use 23 (a.k.a. 15+8) if there’s no DC listed).

The rogue’s goal is to put tension on the lock’s cylinder, and then bind the six pins.

Five of the pins, you bind with d6‘s. Roll them carefully, one at a time, to find out which pins you bind. Keep track of the rolled numbers. The number you roll is the pin you bind. So if you roll a four, that means the fourth pin is binding. If you roll a three, that means the third pin is binding. If you have five dice, you can leave the rolled dice on the table to represent the bound pin.

If you roll an already rolled number, that’s bad! The pin slipped! It’s no longer binding. If you roll that number a third time, it’s binding again. (A fourth time, it slips again, and a fifth time, it’s binding again.)

You can decide to stop rolling early if you’ve found you’ve bound enough pins. It’s not an infinite number of dice, it’s at most five dice.

Once you’ve rolled your five dice, or decided to stop, you’ve got at least one pin left to bind, maybe more if some pins slipped. Roll a d20 and add your normal tool proficiency bonuses as per normal 5e, but also add all the pins you’ve managed to bind with d6‘s. So if you’ve bound the fourth and fifth pins, add nine. Meet or beat the DC to open the lock. If you can’t, the last few pins are beyond your ability and you can’t open the lock.

Math nerds be warned

Ten or to be specific, 4123/432, is the average sum of the pin outcomes when players can stop early. Nine, or to be specific, 1477/162, or approximately 9.12, is the sum when they can’t.

But, it’s a long-tailed curve. If we instead think “What are the chance of the lock opening or not”, it’s better to add eight.

Looking at a variety of lock DCs and modifiers, it stays within 1/20 difference when we add eight.